Finding the error in an argumentChain rule notation for function with two variablesThe multivariable chain rule and functions that depend on themselvesCalculate partial derivative $f'_x, f'_y, f'_z$ where $f(x, y, z) = x^fracyz$Simple Chain Rule for PartialsChain rule for partial derivativesQuestion regarding the proof of the directional derivativePartial derivative of a function w.r.t an argument that occurs multiple timesDerivative of function of matrices using the product ruleWhen to use Partial derivatives and chain rulePartial derivative with dependent variables
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Finding the error in an argument
Chain rule notation for function with two variablesThe multivariable chain rule and functions that depend on themselvesCalculate partial derivative $f'_x, f'_y, f'_z$ where $f(x, y, z) = x^fracyz$Simple Chain Rule for PartialsChain rule for partial derivativesQuestion regarding the proof of the directional derivativePartial derivative of a function w.r.t an argument that occurs multiple timesDerivative of function of matrices using the product ruleWhen to use Partial derivatives and chain rulePartial derivative with dependent variables
$begingroup$
If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$
Therefore
$2xfracpartial zpartial y=0$
and
$fracpartial zpartial y=0$
What is wrong with this argument?
I have a feeling that
1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and
2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.
How is my reasoning? I am pretty confused by this question.
calculus multivariable-calculus partial-derivative
asked 3 hours ago
mathenthusiastmathenthusiast
808
$endgroup$
add a comment |
$begingroup$
If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$
Therefore
$2xfracpartial zpartial y=0$
and
$fracpartial zpartial y=0$
What is wrong with this argument?
I have a feeling that
1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and
2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.
How is my reasoning? I am pretty confused by this question.
calculus multivariable-calculus partial-derivative
asked 3 hours ago
mathenthusiastmathenthusiast
808
$endgroup$
$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
3 hours ago
1
$begingroup$
I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
$endgroup$
– BSplitter
3 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
3 hours ago
add a comment |
$begingroup$
If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$
Therefore
$2xfracpartial zpartial y=0$
and
$fracpartial zpartial y=0$
What is wrong with this argument?
I have a feeling that
1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and
2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.
How is my reasoning? I am pretty confused by this question.
calculus multivariable-calculus partial-derivative
asked 3 hours ago
mathenthusiastmathenthusiast
808
$endgroup$
If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$
Therefore
$2xfracpartial zpartial y=0$
and
$fracpartial zpartial y=0$
What is wrong with this argument?
I have a feeling that
1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and
2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.
How is my reasoning? I am pretty confused by this question.
calculus multivariable-calculus partial-derivative
calculus multivariable-calculus partial-derivative
asked 3 hours ago
mathenthusiastmathenthusiast
808
asked 3 hours ago
mathenthusiastmathenthusiast
808
edited 3 hours ago
mathenthusiast
asked 3 hours ago
mathenthusiastmathenthusiast
808
asked 3 hours ago
mathenthusiastmathenthusiast
808
asked 3 hours ago
mathenthusiastmathenthusiast
808
808
$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
3 hours ago
1
$begingroup$
I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
$endgroup$
– BSplitter
3 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
3 hours ago
add a comment |
$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
3 hours ago
1
$begingroup$
I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
$endgroup$
– BSplitter
3 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
3 hours ago
$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
3 hours ago
$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
3 hours ago
1
1
$begingroup$
I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
$endgroup$
– BSplitter
3 hours ago
$begingroup$
I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
$endgroup$
– BSplitter
3 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
3 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nothing wrong. Just change it into
$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$
Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$
Where I have clearly written down the restriction $y=x^2$.
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Nothing wrong. Just change it into
$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$
Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$
Where I have clearly written down the restriction $y=x^2$.
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
$endgroup$
add a comment |
$begingroup$
Nothing wrong. Just change it into
$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$
Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$
Where I have clearly written down the restriction $y=x^2$.
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
$endgroup$
add a comment |
$begingroup$
Nothing wrong. Just change it into
$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$
Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$
Where I have clearly written down the restriction $y=x^2$.
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
$endgroup$
Nothing wrong. Just change it into
$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$
Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$
Where I have clearly written down the restriction $y=x^2$.
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
answered 3 hours ago
Holding ArthurHolding Arthur
1,370417
1,370417
add a comment |
add a comment |
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$begingroup$
My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
$endgroup$
– BSplitter
3 hours ago
1
$begingroup$
I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
$endgroup$
– BSplitter
3 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
3 hours ago